Check the injectivity and surjectivity of the | vedclass

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(A) The function is $f: N \rightarrow N$ defined by $f(x) = x^{2}$.For injectivity:Let $x, y \in N$ such that $f(x) = f(y)$.Then $x^{2} = y^{2}$.Since

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Injective but not surjective

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(A) The function is $f: N ightarrow N$ defined by $f(x) = x^{2}$.For injectivity:Let $x, y \in N$ such that $f(x) = f(y)$.Then $x^{2} = y^{2}$.Since $x, y \in N$ (natural numbers are positive),we have $x = y$.Therefore,$f$ is injective.For surjectivity:$A$ function is surjective if for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$.Consider $y = 2 \in N$.We need $x^{2} = 2$,which implies $x = \sqrt{2}$.Since $\sqrt{2} 
otin N$,there is no $x \in N$ such that $f(x) = 2$.Therefore,$f$ is not surjective.Conclusion: The function $f$ is injective but not surjective.

Check the injectivity and surjectivity of the following function $f: N \rightarrow N$ defined by $f(x)=x^{2}$.

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